Problem
Let $G$ be a finite group and let $X$ be a $G$-set. Assume that $|G|$ does not divide $|X|$. Prove that there exists a $g \in G$ and an $x\in X$ such that $g \neq e$ and such that $gx = x$.
Hint: Use Burnside's formula
Solution:
We are given that $|G|$ does not divide $|X|$. This means that $|X|$ cannot be written as a multiple of $|G|$. If $|$Fix$(g)| = |X|$ for all elements $g \in G$, then the number of orbits of $X$ would be equal to $1$, because every element of $X$ would be fixed by every element of $G$. But this would imply that $|X| = |G|$, which is a contradiction. So there must be at least one element $g \in G$ that fixes fewer than $|X|$ elements. This means that there must be at least one element $x \in X$ that is not fixed by $g$, that is, $gx \neq x$. Since $g \neq e$ (the identity element of $G$), this element $x$ satisfies the desired conditions.
By contradiction, let all the pointwise stabilizers be trivial. Then, all the orbits have size $|G|$ (orbit-stabilizer), and hence $|X|$ is a multiple of $|G|$, namely $|G|$ divides $|X|$: contradiction.
For a proof by Burnside's (counting) lemma, recall that $\sum_{g\in G}\left|\operatorname{Fix}(g)\right|=$ $\sum_{x\in X}\left|\operatorname{Stab}(x)\right|$, so if all the stabilizers are trivial, then the number of orbits is given by $\frac{|X|}{|G|}$, which is not an integer if $|G|$ does not divide $|X|$. Or, more straighforwardly, note that from the negation of your condition follow $\operatorname{Fix}(e)=X$ and $\operatorname{Fix}(g)=\emptyset$ otherwise, whence back to the previous contradiction.